ong>#steps⬇️⬇️

1. draw a ray OA with intial point O.
2. Taking O as a centre and some radius, draw an arc of a CIRCLE, which intersects OA, at a point B.
3. Taking B as centre and with the same radius as before, draw an arc intersecting the previous drawn arc, at a point C.
4. Taking C as centre and with the same radius as before, draw an arc intersecting the arc drawn in step 2, say at D.
5. Draw the ray OE passing through C. Then ∠EOA = 60∘ .
6. Draw the ray OF passing through D. Then ∠ FOE = 60∘ .
7. Next, taking C and D as centres and with radius more than ½ CD, draw arcs to intersect each other, at G.
8. Draw the ray OG, which is the bisector of the angle FOE, i.e., ∠FOG = ∠EOG = 1/2 ∠FOE = 1/2 (60∘ ) = 30∘ .
9. Thus, ∠GOA = ∠GOE + ∠ EOA = 30∘ + 60∘ = 90∘.Now taking O as centre and any radius more than OB, draw an arc to intersect the rays OA and OG, at H and I.
10. Next, taking H and I as centres and with the radius more than 1/2 HI, draw arcs to ntersect each other, at J.
11. Draw the ray OJ. This ray OJ is the required bisector of the ∠ GOA. Thus, ∠GOJ = ∠AOJ = 1/2 ∠GOA = 1/2(90∘) = 45∘.

#Justification:

(i)Join BC.

Then, OC = OB = BC triangle. (By construction)

∴ ∠COB is an equilateral triangle.

∴ ∠COB = 60∘.

∴ ∠EOA = 60∘.

(ii)Join CD.

Then, OD = OC = CD (By construction)

∆DOC is an equilateral triangle.

∴ ∠DOC = 60∘.

∴ ∠ FOE = 60∘.

(iii)Join CG and DG.

In ΔODG and ΔOCG,

OD = OC[ Radii of the same arc]

DG=CG [Arcs of equal radii]

OG=OG [Common]

∴ Δ ODG = ΔOCG [SSS Rule]

∴ ∠ DOG= ∠COG [CPCT]

∴ ∠FOG = ∠ EOG = 1/2 ∠FOE = 1/2 (60∘) = 30∘

Thus, ∠GOA = ∠GOE + ∠EOA = 30∘ + 60∘ = 90∘

∴ ∠AOJ= ∠GOJ= 1/2 ∠GOA = ½(90°)=45

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